K2SO4 H3PO4 and NaOH KOH and H2SO4 HPO4 and KOH

Give evidence to support or dispute: “In nature, the chance of finding one isotope of an element is the same for all elements.”

alekssr [168]

Answer:

False, isotopes have different occurrence percentages, so the changes are different.

Explanation:

Hello,

In this case, since it is false that the isotopes of all the elements can be found with the same chance (occurrence) we can consider the following facts:

1. Carbon atom has two major occurring isotopes: C-12 (98.93%) and C-13 (1.07%).

2. Bromine atom has two major occurring isotopes: Br-79 (50.69%) and Br-81 (49.31%).

3. Calcium has four major occurring isotopes: Ca-40 (96.941%), Ca-42 (0.647%), Ca-43 (0.135%) and Ca-44 (2.086%).

Which show us that the chances of finding any isotope differ among elements.

Regards.

5 0

3 years ago

How does the electron configuration of elements within the same group compare? (A.) They all have their valence electrons in the

BaLLatris [955]

Answer:

( B) They all have their valence electrons in the same type of subshell.

Explanation:

With each period, a new shell is added to the atom.

Further, the groups are classified based on the type of subshell the last electron enters and number of valence electrons.

For all elements of same group, the last electron enters the same type of subshell.

Say, for group 1, last electron enters s orbital and they have 1 valence elctron.

for group 17, last electron enters p orbital and they have 7 valence electrons.

(A) and (D) are wrong because, energy level of the valence electrons is determined by the principle quantum number n and l and not by the type of subshell(only l) they enter.

(C) if the valence electron enters p orbital, then the elements will be placed in the p- block.

3 0

3 years ago

1. What is the relationship between the concentration of the hydronium and hydroxide

Hitman42 [59]

Answer:

- Neutral solutions: concentration of hydronium equals the concentration of hydroxide.

- Acid solutions: concentration of hydronium is greater than the concentration of hydroxide.

- Basic solutions concentration of hydronium is lower than the concentration of hydroxide.

Explanation:

Hello,

It is widely known that the pH of water is 7, therefore the pOH of water is also 7 based on:

$pH+pOH=14$

In such a way, we can compute the concentration of hydronium and hydroxide ions as shown below:

$pH=-log([H^+])\\$

$[H^+]=10^{-pH}=10^{-7}=1x10^{-7}M$

$pOH=-log([OH^-])$

$[OH^-]=10^{-pOH}=10^{-7}=1x10^{-7}M$

Thus, we notice that the relationship between the concentration of the hydronium is equal for water or neutral solutions. Moreover, if we talk about acid solutions, pH<OH therefore the concentration of hydronium is greater than the concentration of hydroxide. On the other hand if we talk about basic solutions, pH>OH therefore the concentration of hydronium is lower than the concentration of hydroxide.

Best regards.

4 0

3 years ago

The moon will always appear over the eastern horizon at the same time each night. true or false

Otrada [13]

False it will be false

7 0

3 years ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$